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Q.
The region represented by the solution set of the system of inequalities $x-y \leq 1, \quad x+2 y \leq 8$, $2 x+y \geq 2, x \geq 0, y \geq 0$ is
Linear Inequalities
Solution:
Given inequalities are
$x-y \leq 1 $...(i)
$x+2 y \leq 8 $...(ii)
$2 x+y \geq 2$...(iii)
$ x \geq 0 $...(iv)
$ y \geq 0$....(v)
The lines corresponding to (i), (ii) and (iii) are
$x-y=1 $...(vi)
$x+2 y=8 $...(vii)
$2 x+y=2$....(viii)
Draw the graph of line (vi), (vii) and (viii).
$\because 0-0 \leq 1$ i.e., $0 \leq 1$, which is true.
Therefore, inequality (i) represent the half plane made by the line (vi), which contains origin.
Now, $0+2(0) \leq 8$ i.e., $0 \leq 8$, which is true. So, inequality (ii) represent the half plane made by line (vii) which contain origin.
Again, $2(0)+0 \geq 2$ i.e., $0 \geq 2$, which is not true.
$\therefore$ Inequality (iii) represent the half plane made by the line (viii), which does not contain origin.
Also, $x \geq 0, y \geq 0$ represent the region in I quadrant.
The region common to (i), (ii), (iii), (iv) and (v) is the shaded region which is bounded.
